Given that,

D_{1} = 60cm

D_{2} = 24cm

N_{2 }= 300rpm µ = 0.3

X = 3m = 300cm

P = 3.75KW

Safe Tension = Maximum tension = 100N/cm width = 100b N b = width of belt

T_{max} = 100b ...(i)

Let θ = Angle of contact

Sinα = (r_{1} – r_{2})/X = (30 – 12)/300 ; α = 3.45°, ...(ii)

= Π – 2α = (180 – 2 X 3.45) = 173.1°

= (173.1°)X Π/180 = 3.02 rad ......(iii)

Now,

Using the relation, Ratio of belt tension = T_{1}/T_{2} = e^{µθ} = e^{(0.3)(3.02)}

T_{1} = 2.474 T_{2} ......(iv)

Now,

V = πDN/60 m/sec, D is in meter and N is in RPM

= 3.14 X (0.24)(300)/60 = 3.77m/sec ......(v)

Power Transmitted (P) = (T_{1} – T_{2}).v/1000 Kw

3.75 = (T_{1} – T_{2})X 3.77/1000

T_{1} – T_{2} = 994.7N .......(vi)

From relation (iv) and (v), we get:

T_{1} = 1669.5N ...........(vii)

T_{2 }= 674.8N ...........(viii)

(i) For width of the belt

But T_{1} = T_{max }= 100b; 1669.5 = 100b; b = 16.7cm

(ii) For initial tension in the belt

Let To = initial tension in the belt

To = (T_{1} + T_{2})/2

= (1669.5 + 674.8)/2

**To = 1172.15N**

(iii) For length of belt

Putting all the value, we get

**L = 7.33m.**